Capacitor discharge probes

Hi Shaun,

I could say a few words here but Sam can provide you with much more info. See here:

Tom

Thanks for the link, there is alot of information there but not what I was asking for.

How do you calculate resistor power rating required to discharge a capacitor of x volts and y microfarads?

Shaun

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The vast majority of capacitors will be discharged by the time you get your test probes ready to use. I've been doing ESR stuff several years and only once did I damage a meter by sending power into a Capacitor Wizard. It was a small value (around 10 ohms) resistor that fried and only took a few minutes to replace. Or do you just want formulas?

G=B2

That's a bit of a weird question. Most times caps aren't charged (much) just sitting on the bench. So any old resistor will do. (Is there some danger just using a peice of wire?)

The energy stored in a cap is 1/2 CV^2. If you discharge it with a resistor R it'll take something like the time constant RC. So the power is energy/time ~ V^2/R. (Hmm the C goes away?) anyway, if the RC time is short the resistor won't have time to heat up much and you could use a smaller value. So what's a short time? at a guess anything shorter than 1 s or maybe 0.1 s.

George H.

"Shaun"

** You are labouring under a mis-apprehension.

Any resistor connected across a capacitors will EVENTUALLY discharge it.

The only things to worry abut are how long will it take and might the resistor become damaged by too much current .

A 100 ohm, 5W *wire wound* resistor will discharge almost any electro ( to a safe voltage) in no more than a few seconds - however a very large, high voltage electro might have enough energy to destroy it.

.... Phil

The vast majority of capacitors will be discharged by the time you get your test probes ready to use. I've been doing ESR stuff several years and only once did I damage a meter by sending power into a Capacitor Wizard. It was a small value (around 10 ohms) resistor that fried and only took a few minutes to replace. Or do you just want formulas?

** The " Blue ESR Meter " will be damaged by voltages more than about 50 volts DC

..... Phil

A 1K resistor will discharge a 1000uF capacitor in 5s. A 100R resistor achieves the same in 0.5s. Time constant is tau=R·C and in 4 or 5 tau you have discharged your capacitor.

The power absorbed by the resistor at the moment of closing the circuit is V²/R. For V=63V and R=100R, this is roughly 40W. At t=0.1s voltage is just 23V and instantaneous power absorbed 5W. Your jig will certainly survive such short spikes. If you take 1K, a single resistor will achieve the same at the cost of some extra time :)

Pere

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** All perfectly correct.

However, a physically small resistor can only absorb so much energy in a short space of time before the conductor inside MELTS !!

The energy stored in a capacitor is give by the formula:

E = 0.5C x V squared.

Taking a practical, worst case example of a 1000uF cap charged to 400 volts - the stored energy is 80 joules, most of which is dumped in the first 0.2 seconds.

Depends very much on the construction of the particular resistor whether it can absorb such a large energy spike or not.

The best type is carbon composition, then wire wound and last of all deposited film resistors.

.... Phil

Just made a quick search on some Farnell 100R resistors. It is interesting to see that a 1W CCR resistor such as CCR1100RKTB (Tyco) has a pulse limiting power of 600W for single 0.1s pulses.

Pere

I'd guess that a resistor can easily absorb E joules, where E = P * T and P is its power rating and T is its thermal time constant. T is maybe 10 or more seconds for something like a carbon comp, maybe less for a wirewound or axial carbon film, a fraction of a second for a surface-mount thickfilm.

So it would probably be OK for a 2 watt axial resistor to absorb 20 joules. I've seen specs for 5-watt pulse-rated resistors that are claimed to absorb over 100 joules.

Somebody should destroy some resistors and publish the results. I exploded some 7.5 ohm 1206 thickfilm resistors. It takes about 60 watts in 1 millisecond. Pulsed, they look like LEDs in the infrared.

John

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How big a carbon comp? I would have guessed something less.. maybe 1 second. Hmm, just wondering if you can really define a thermal time constant for a resistor. At least for a carbon comp, the heat is being generated everywhere inside the thing. I think of a thermal time constant as heating one end of something and asking how long it takes the whole thing to warm up.

Perhaps it's the heat capacity that determines the maximum energy it can dissipate. E =3D C delta T.

George H.

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Most circuitry with powered capacitors has a bleed resistor to discharge those capacitors in a few seconds. For safety, one is advised not to trust the bleed resistor, and various kinds of probes up to 'shorting chains' and down to 'connect with ammeter setting on your VOM' can be used to ensure discharge.

For 15 kV capacitors in one-meter-cube sizes, use a steel (but not galvanized) chain on a long insulating pole, with eye protection. For PC power supply filter capacitors, try that 100 ohm 5W resistor (use a carbon resistor if you can, because the initial surge will be more than 5W).

Did we all just witness a lovefest from Phil? :-) Mike

Thanks for all your replies.

So I guess the important factor is it's pulse power rating for a resistor, although I haven't seen any resistors with this spec before. If I work out the math now that I know what to do thanks to you guys, P = 1/10 V^2/R for 5 tau. What confuses me is that C cancels out, so it wouldn't matter what value of C I'm discharging, the only thing that changes is the amount of time it takes to discharge.

Does anyone know a rule of thumb for pulse power verses continuous power for a given time frame for a resistor??

thanks again,

Shaun

It's interesting how many people speculate and simulate and don't actually measure stuff.

ftp://jjlarkin.lmi.net/A-B_tau.gif

Tau is roughly 40 seconds.

At 10 watts, this resistor gets too hot to hold (my threshold is about

55C) in around 10 seconds. That's 100 joules.

So I'd guess a 2-watt carbon comp can easily absorb 100 joules, probably a lot more.

John

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This is sitting in air? Aren't you measuring the convective cooling? I bet the slope at the beginning gives the heat capacity.

I'm not sure what you mean by tau? At ten watts didn't it heat up twice as fast?

Perhaps we have different definitions for thermal time constant (TC). I think of measuring TC by having a piece of stuff and a heat source. I pulse the heat source and measure how long it takes the hunk of stuff to come to a new temperature. If the heat pulse is twice as big the temperature rise is twice as large, but it takes the same amount of time. (Well as long as everything stays linear.)

I did try and measure (sometime last year) how much of the heat comes out of the leads of a resistor and how much comes out through the body. These were cheap Xicon 1/4 watt metal film resistors. I thought I'd see more than 1/2 the heat leaving from the leads. But I found that most of it was escaping though the body. A collegaue(sp) told me my measurements were crap once the temperature got above a few degrees cause then convection takes over. (Oh I was really trying to answer the converse question, is the temperature of the resistor determned by the temperature of it's leads or the temperature of the gas that surrounds the body.)

George H.

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OK If I'm reminded of it I'll try that on Monday. 100 Watts into a

2watt carbon comp for 1 second. The only easy way I can think to do this is by taking a 35V 3A supply and ramping up the current knob to max for one second...or maybe pushing the power button on and off. (hmm I could build some power fet switch thing, but that sounds like work.) I hope I have some 2Watt 10 ohm carbon comps. If I only find 1/2 watt ones can I scale this down by a factor of four? 25 watts for 1 second. Hey anyone with a home lab feel like trying to blow things up this weekend.

George H.

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Yes. At the end of the curve, the applied power is almost balanced by heat losses.

Yes.

Tau is the time it takes my graph to go from start to 63% of its terminal value. It's the thermal time constant of the resistor.

Sure, the rate of temperature rise is proportional to the applied power. More watts will zoom my graph vertically. Tau won't change, at least until something melts, or convection gets nonlinear maybe.

This graph does suggest how many joules the resistor could absorb before it hits some scary temperature.

John

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Opps my mistake.. twice the heat, twice the slope, but twice the final temp difference so same TC. (late night mistake... ah heck I make mistakes all day long.)

Seems like if you could estimate the temperature at which the resistor 'craps out' then you could use the slope of your curve at the origin to guess at the total energy. (Sorry this is perhaps obvious to you, I'm just a bit slow.)

George H.